An investments value, V(t) is modelled by the function V(t)=2500(1.15)^2, where t is the number of years after funds are invested
A) find the instantaneous rate of change in the value of the investment at t=4, what intervals would you choose? Why?
My question is ... Which intervals should I use? And is there a way I could do it using the limit x->h (f(a+h)-f(a))/a method?
instaneous rate= dV/dt
= limx->h (f(a+h)-f(a))/a
Intervals have to be t greater than zero. Why? what does negative times mean?
Now the issue I have is the V(t) function itself: IT IS NOT A FUNCTION OF t, nowhere does t appear in the function. You either typed it incorrectly, or your teacher is having a serious senior moment.
To find the instantaneous rate of change in the value of the investment at t=4 using the limit definition of derivative, you can use the formula:
f'(a) = lim(h->0)[(f(a+h) - f(a))/h]
In this case, a represents the value of t at which you want to find the instantaneous rate of change (t=4).
Let's calculate it step-by-step:
1. Substitute the given function V(t) into the formula:
V'(4) = lim(h->0)[(V(4+h) - V(4))/h]
2. Substitute the given values into the equation:
V'(4) = lim(h->0)[(2500(1.15)^(4+h) - 2500(1.15)^4)/h]
3. Simplify the expression:
V'(4) = lim(h->0)[2500(1.15)^(4+h) - 2500(1.15)^4]/h
4. Apply the limit definition of derivative by evaluating the limit as h approaches 0:
V'(4) = lim(h->0)[2500(1.15)^(4+h) - 2500(1.15)^4]/h
Now, to evaluate this limit, you can use numerical methods or a calculator capable of handling limits, such as a graphing calculator or an online limit calculator. By inputting the expression into one of these tools, you can find the instantaneous rate of change in the value of the investment at t=4.
Regarding the intervals, it's not specified in the question which intervals you need to use. However, if you want to find the instantaneous rate of change at specific values of t (such as t=4), it's recommended to use infinitesimally small intervals around the desired point.
To find the instantaneous rate of change in the value of the investment at t = 4, you need to calculate the derivative of the function V(t) with respect to t. In this case, the function is given as V(t) = 2500(1.15)^2.
Using the power rule for differentiation, if f(t) = a * b^t, then f'(t) = a * ln(b) * b^t. Applying this to the given function, we have:
V'(t) = 2500 * ln(1.15) * (1.15)^t
Now, to find the instantaneous rate of change at t = 4, you can substitute t = 4 into the derivative function.
V'(4) = 2500 * ln(1.15) * (1.15)^4
Computing this value will give you the instantaneous rate of change in the value of the investment at t = 4.
Regarding the method of using the limit definition of the derivative, which involves evaluating the expression (f(a + h) - f(a))/h as h approaches zero, you could apply it as follows:
Start with the expression:
[f(a + h) - f(a)]/h
In this case, a = 4 and f(t) = 2500(1.15)^2. Substituting these values, we have:
[f(4 + h) - f(4)]/h = [2500(1.15)^(4 + h) - 2500(1.15)^2]/h
Finally, as h approaches zero, evaluate the expression to find the instantaneous rate of change in the value of the investment at t = 4.
However, using the power rule to differentiate the expression V(t) = 2500(1.15)^2 is a simpler and more efficient method in this case.